You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array `nums` representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

```Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.
```

Example 2:

```Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.
```

Constraints:

• `1 <= nums.length <= 100`
• `0 <= nums[i] <= 400`

Idea:

Dynamic Programing

Try to look for a pattern, see comments in the code.

Solution:

``````/**
* @param {number[]} nums
* @return {number}
*/
var rob = function(nums) {
// one house
if (nums.length === 1) return nums[0];

// dp[i] = max money can be taken after visited i houses
let dp = new Array(nums.length);
dp[1] = nums[0];
dp[2] = Math.max(nums[0], nums[1]);
// two houses
if (nums.length === 2) return dp[2];

// more houses - try to look for a pattern
// dp[3] = Math.max(dp[2], dp[1] + nums[2]);
// dp[4] = Math.max(dp[3], dp[2] + nums[3]);
// ...
for (let i = 3; i <= nums.length; i++) {
dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);
}
return dp[nums.length];
};``````